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Math  /  Algebra

QuestionAvery and Collin were trying to challenge each other with equations for sequences. Avery was looking at an explicit equation that Collin wrote. t(n)=4.5n−8t(n)=4.5 n-8 a. Write the first 4 terms for the sequence. b. What would Avery do to write the 15th 15^{\text {th }} term of this sequence? c. Write a recursive equation for this sequence.

Studdy Solution

STEP 1

1. The explicit equation for the sequence is given as t(n)=4.5n−8 t(n) = 4.5n - 8 .
2. We need to find specific terms in the sequence using this equation.
3. A recursive equation expresses each term based on the previous term(s).

STEP 2

1. Calculate the first 4 terms of the sequence using the explicit formula.
2. Use the explicit formula to find the 15th term.
3. Derive a recursive equation for the sequence.

STEP 3

To find the first term t(1) t(1) , substitute n=1 n = 1 into the explicit equation:
t(1)=4.5(1)−8 t(1) = 4.5(1) - 8 t(1)=4.5−8 t(1) = 4.5 - 8 t(1)=−3.5 t(1) = -3.5

STEP 4

To find the second term t(2) t(2) , substitute n=2 n = 2 into the explicit equation:
t(2)=4.5(2)−8 t(2) = 4.5(2) - 8 t(2)=9−8 t(2) = 9 - 8 t(2)=1 t(2) = 1

STEP 5

To find the third term t(3) t(3) , substitute n=3 n = 3 into the explicit equation:
t(3)=4.5(3)−8 t(3) = 4.5(3) - 8 t(3)=13.5−8 t(3) = 13.5 - 8 t(3)=5.5 t(3) = 5.5

STEP 6

To find the fourth term t(4) t(4) , substitute n=4 n = 4 into the explicit equation:
t(4)=4.5(4)−8 t(4) = 4.5(4) - 8 t(4)=18−8 t(4) = 18 - 8 t(4)=10 t(4) = 10
The first four terms of the sequence are: −3.5,1,5.5,10 -3.5, 1, 5.5, 10 .

STEP 7

To find the 15th term t(15) t(15) , substitute n=15 n = 15 into the explicit equation:
t(15)=4.5(15)−8 t(15) = 4.5(15) - 8 t(15)=67.5−8 t(15) = 67.5 - 8 t(15)=59.5 t(15) = 59.5

STEP 8

To write a recursive equation, we need to express t(n) t(n) in terms of t(n−1) t(n-1) . From the explicit formula, the difference between consecutive terms is constant:
t(n)−t(n−1)=4.5n−8−(4.5(n−1)−8) t(n) - t(n-1) = 4.5n - 8 - (4.5(n-1) - 8) t(n)−t(n−1)=4.5n−8−4.5n+4.5−8 t(n) - t(n-1) = 4.5n - 8 - 4.5n + 4.5 - 8 t(n)−t(n−1)=4.5 t(n) - t(n-1) = 4.5
Thus, the recursive formula is: t(n)=t(n−1)+4.5 t(n) = t(n-1) + 4.5 with the initial term t(1)=−3.5 t(1) = -3.5 .
The first four terms are: −3.5,1,5.5,10 -3.5, 1, 5.5, 10 . The 15th term is: 59.5 59.5 . The recursive equation is: t(n)=t(n−1)+4.5 t(n) = t(n-1) + 4.5 , with t(1)=−3.5 t(1) = -3.5 .

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